Question: $\dfrac{d}{dx}\left[\dfrac{x^3}{\sin(x)}\right]=$
Solution: $\dfrac{x^3}{\sin(x)}$ is the quotient of two, more basic, expressions: $x^3$ and $\sin(x)$. Therefore, the derivative of the expression can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[\dfrac{x^3}{\sin(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}[x^3]\sin(x)-x^3\dfrac{d}{dx}[\sin(x)]}{[\sin(x)]^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{3x^2\sin(x)-x^3\cos(x)}{\sin^2(x)}&&\gray{\text{Differentiate }x^3\text{ and }\sin(x)} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[\dfrac{x^3}{\sin(x)}\right]=\dfrac{3x^2\sin(x)-x^3\cos(x)}{\sin^2(x)}$ or any other equivalent form.